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Informed search algorithms

Chapter 4

Outline

  Best-first search   Greedy best-first search   A* search   Heuristics   Memory Bounded A* Search

Best-first search   Idea: use an evaluation function f(n) for each node

  f(n) provides an estimate for the total cost.   Expand the node n with smallest f(n).

  Implementation: Order the nodes in fringe increasing order of cost.

  Special cases:   greedy best-first search   A* search

Romania with straight-line dist.

Greedy best-first search

  f(n) = estimate of cost from n to goal   e.g., f(n) = straight-line distance from n

to Bucharest   Greedy best-first search expands the

node that appears to be closest to goal.

Greedy best-first search example

Greedy best-first search example

Greedy best-first search example

Greedy best-first search example

Properties of greedy best-first search

  Complete? No – can get stuck in loops.   Time? O(bm), but a good heuristic can

give dramatic improvement   Space? O(bm) - keeps all nodes in

memory   Optimal? No e.g. AradSibiuRimnicu

VireaPitestiBucharest is shorter!

Think of an

example

Properties of greedy best-first search

gc

b

a

d

start state

goal state

f(n) = straightline distance

A* search

  Idea: avoid expanding paths that are already expensive

  Evaluation function f(n) = g(n) + h(n)   g(n) = cost so far to reach n   h(n) = estimated cost from n to goal   f(n) = estimated total cost of path through n

to goal   Best First search has f(n)=h(n)   Uniform Cost search has f(n)=g(n)

Admissible heuristics

  A heuristic h(n) is admissible if for every node n, h(n) ≤ h*(n), where h*(n) is the true cost to reach the goal state from n.

  An admissible heuristic never overestimates the cost to reach the goal, i.e., it is optimistic

  Example: hSLD(n) (never overestimates the actual road distance)

  Theorem: If h(n) is admissible, A* using TREE-SEARCH is optimal

Admissible heuristics E.g., for the 8-puzzle:   h1(n) = number of misplaced tiles   h2(n) = total Manhattan distance (i.e., no. of squares from desired location of each tile)

  h1(S) = ?   h2(S) = ?

Admissible heuristics E.g., for the 8-puzzle:   h1(n) = number of misplaced tiles   h2(n) = total Manhattan distance (i.e., no. of squares from desired location of each tile)

  h1(S) = ? 8   h2(S) = ? 3+1+2+2+2+3+3+2 = 18

Dominance   If h2(n) ≥ h1(n) for all n (both admissible)   then h2 dominates h1   h2 is better for search: it is guaranteed to expand less or equal nr of nodes.

  Typical search costs (average number of nodes expanded):

  d=12 IDS = 3,644,035 nodes A*(h1) = 227 nodes A*(h2) = 73 nodes

  d=24 IDS = too many nodes A*(h1) = 39,135 nodes A*(h2) = 1,641 nodes

Relaxed problems   A problem with fewer restrictions on the actions

is called a relaxed problem   The cost of an optimal solution to a relaxed

problem is an admissible heuristic for the original problem

  If the rules of the 8-puzzle are relaxed so that a tile can move anywhere, then h1(n) gives the shortest solution

  If the rules are relaxed so that a tile can move to any adjacent square, then h2(n) gives the shortest solution

Consistent heuristics   A heuristic is consistent if for every node n, every successor n'

of n generated by any action a,

h(n) ≤ c(n,a,n') + h(n')

  If h is consistent, we have

f(n’) = g(n’) + h(n’) (by def.) = g(n) + c(n,a,n') + h(n’) (g(n’)=g(n)+c(n.a.n’)) ≥ g(n) + h(n) = f(n) (consistency) f(n’) ≥ f(n)

  i.e., f(n) is non-decreasing along any path.

  Theorem: If h(n) is consistent, A* using GRAPH-SEARCH is optimal

It’s the triangle inequality !

keeps all checked nodes in memory to avoid repeated states

A* search example

A* search example

A* search example

A* search example

A* search example

A* search example

Properties of A*   Complete? Yes (unless there are infinitely many

nodes with f ≤ f(G) , i.e. step-cost > ε)   Time/Space? Exponential except if:   Optimal? Yes   Optimally Efficient: Yes (no algorithm with the same heuristic is guaranteed to expand fewer nodes)

Optimality of A* (proof)   Suppose some suboptimal goal G2 has been generated and is in

the fringe. Let n be an unexpanded node in the fringe such that n is on a shortest path to an optimal goal G.

  f(G2) = g(G2) since h(G2) = 0   f(G) = g(G) since h(G) = 0   g(G2) > g(G) since G2 is suboptimal

  f(G2) > f(G) from above   h(n) ≤ h*(n) since h is admissible (under-estimate)   g(n) + h(n) ≤ g(n) + h*(n) from above   f(n) ≤ f(G) since g(n)+h(n)=f(n) & g(n)+h*(n)=f(G)   f(n) < f(G2) from

We want to prove: f(n) < f(G2) (then A* will prefer n over G2)

SEARCH TREE

1) Consider the search tree to the right. There are 2 goal states, G1 and G2. The numbers on the edges represent step-costs. You also know the following heuristic estimates: h(BG2) = 9, h(DG2)=10, h(AG1)=2, h(CG1)=1

a) In what order will A* search visit the nodes? Explain your answer by indicating the value of the evaluation function for those nodes that the algorithm considers.

G1 G2

C D

A B

R 1

2

10

9

1

1

Exercise

S B

A D

E

C

F

G

1 20

2

3

4 8

6 1 1

straight-line distances

h(S-G)=10 h(A-G)=7 h(D-G)=1 h(F-G)=1 h(B-G)=10 h(E-G)=8 h(C-G)=20

The graph above shows the step-costs for different paths going from the start (S) to the goal (G). On the right you find the straight-line distances.

1.  Draw the search tree for this problem. Avoid repeated states.

2.  Give the order in which the tree is searched (e.g. S-C-B...-G) for A* search. Use the straight-line dist. as a heuristic function, i.e. h=SLD,

and indicate for each node visited what the value for the evaluation function, f, is.

try yourself

Memory Bounded Heuristic Search: Recursive BFS

  How can we solve the memory problem for A* search?

  Idea: Try something like depth first search, but let’s not forget everything about the branches we have partially explored.

  We remember the best f-value we have found so far in the branch we are deleting.

RBFS:

RBFS changes its mind very often in practice.

This is because the f=g+h become more accurate (less optimistic) as we approach the goal. Hence, higher level nodes have smaller f-values and will be explored first.

Problem: We should keep in memory whatever we can.

best alternative over fringe nodes, which are not children: i.e. do I want to back up?

Simple Memory Bounded A*

  This is like A*, but when memory is full we delete the worst node (largest f-value).

  Like RBFS, we remember the best descendent in the branch we delete.

  If there is a tie (equal f-values) we delete the oldest nodes first.

  simple-MBA* finds the optimal reachable solution given the memory constraint.

  Time can still be exponential. A Solution is not reachable if a single path from root to goal does not fit into memory

SMA* pseudocode (not in 2nd edition 2 of book)

function SMA*(problem) returns a solution sequence inputs: problem, a problem static: Queue, a queue of nodes ordered by f-cost

Queue MAKE-QUEUE({MAKE-NODE(INITIAL-STATE[problem])}) loop do if Queue is empty then return failure n deepest least-f-cost node in Queue if GOAL-TEST(n) then return success s NEXT-SUCCESSOR(n) if s is not a goal and is at maximum depth then f(s) ∞ else f(s) MAX(f(n),g(s)+h(s)) if all of n’s successors have been generated then update n’s f-cost and those of its ancestors if necessary if SUCCESSORS(n) all in memory then remove n from Queue if memory is full then delete shallowest, highest-f-cost node in Queue remove it from its parent’s successor list insert its parent on Queue if necessary insert s in Queue end

Simple Memory-bounded A* (SMA*)

24+0=24

A

B G

C D

E F

H

J

I

K

0+12=12

10+5=15

20+5=25

30+5=35

20+0=20

30+0=30

8+5=13

16+2=18

24+0=24 24+5=29

10 8

10 10

10 10

8 16

8 8

f = g+h

(Example with 3-node memory) Progress of SMA*. Each node is labeled with its current f-cost. Values in parentheses show the value of the best forgotten descendant.

Algorithm can tell you when best solution found within memory constraint is optimal or not.

☐ = goal Search space

maximal depth is 3, since memory limit is 3. This branch is now useless.

best forgotten node

A 12

A

B

12

15

A

B G

13

15 13 H

13

∞

A

G

18

13[15]

A

G 24[∞]

I

15[15]

24

A

B G

15

15 24 ∞

A

B

C

15[24]

15

25

A

B

D

8

20

20[24]

20[∞]

best estimated solution so far for that node

Conclusions

  The Memory Bounded A* Search is the best of the search algorithms we have seen so far. It uses all its memory to avoid double work and uses smart heuristics to first descend into promising branches of the search-tree.